Question: A secant line intersects the curve $y=\ln(x)$ at two points with $x$ -coordinates $2$ and $2+h$. What is the slope of the secant line? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\ln(2+h)-\ln(2)}{2+h}$ (Choice B) B $\dfrac{\ln(2+h)-\ln(2)}{2}$ (Choice C) C $\dfrac{\ln(2+h)-\ln(2)}{h}$ (Choice D) D $\dfrac{\ln(h)-\ln(2)}{h}$
We are given that the secant line intersects the curve at $x=2$ and $x=2+h$. Since these points are on the curve $y=\ln(x)$, we know that their $y$ -values are $y=\ln(2)$ and $y=\ln(2+h)$, correspondingly. To summarize this part, we know that the secant line passes through the points $(2,\ln(2))$ and $(2+h,\ln(2+h))$. This should be enough to find the slope of that line. $\begin{aligned} \text{Slope}&=\dfrac{\text{Change in }y}{\text{Change in }x} \\\\ &=\dfrac{\ln(2+h)-\ln(2)}{2+h-2} \\\\ &=\dfrac{\ln(2+h)-\ln(2)}{h} \end{aligned}$ In conclusion, the slope of the secant line is $\dfrac{\ln(2+h)-\ln(2)}{h}$.